Revision notes · Topic 21 of 25 · A2

Alternating currents

Sinusoidal currents, r.m.s. values, and the rectifying and smoothing of a.c.

Syllabus 21.1 to 21.2 Tier A Level · A2 Prepared by the TheLucidSTEM team

§ 21.1 Characteristics of alternating currents

Key ideas
  • A sinusoidal alternating current is x = x₀ sin ωt, with peak value x₀, period T and frequency f.
  • Over a cycle the mean power is half the peak power, because sin² averages to a half.
  • The root-mean-square value is the steady d.c. that would deliver the same mean power.
Equations
I_rms = I₀ / √2also V_rms = V₀ / √2 for a sinusoidA, V
⟨P⟩ = ½ I₀ V₀mean power = half the peak powerW
I t I₀ peak I_rms = I₀/√2
Fig. 1 · A sinusoidal current: the peak is I₀, while the r.m.s. value (about 0.71 I₀) is the equivalent steady current for power.
Watch out: mains "230 V" is the r.m.s. value; its peak is 230 × √2 ≈ 325 V. Use r.m.s. values for power, but the peak for insulation and breakdown.

§ 21.2 Rectification and smoothing

Key ideas
  • Half-wave rectification (one diode) passes only the positive half-cycles; full-wave rectification (a four-diode bridge) flips the negative halves up too.
  • A smoothing capacitor in parallel with the load charges at each peak and discharges between peaks, reducing the ripple.
  • A larger C or R (a larger time constant) gives a smoother output with smaller ripple.
V t smoothed output (ripple)
Fig. 2 · Full-wave rectified output (solid humps) with a smoothing capacitor: the dashed line shows the capacitor holding the voltage up between peaks, leaving only a small ripple.
Watch out: full-wave rectification gives a ripple at twice the supply frequency, because both halves of each cycle now produce a hump. A 50 Hz supply gives 100 humps per second.